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9 Ray Optics And Optical Instruments
Introduction
The human eye can perceive a narrow range of electromagnetic radiation, which is known as light. This visible light spectrum typically spans wavelengths from approximately 400 nm to 750 nm.
Based on everyday observations, we understand two key properties of light: its incredibly high speed and its tendency to travel in straight lines. While the straight-line travel is an intuitive concept, the finite nature and immense speed of light were discoveries made over time. The accepted speed of light in a vacuum is $c \approx 3 \times 10^8 \text{ m s}^{-1}$, representing the maximum speed attainable in nature.
The concept of light traveling in straight lines appears to contrast with its nature as an electromagnetic wave. This apparent contradiction is resolved by considering the relative sizes of light wavelengths and the objects it interacts with. When light's wavelength is significantly smaller than the dimensions of objects it encounters (which is true for visible light and everyday objects), the wave propagation can be effectively approximated as travel along a straight path. This straight path is termed a ray of light, and a collection of such rays forms a beam of light.
This chapter explores various optical phenomena, including reflection, refraction, and dispersion, by employing this ray picture of light. We will apply the fundamental laws of reflection and refraction to understand how images are formed by both flat and curved mirrors and lenses. Furthermore, we will investigate the construction and operation of common optical devices, including the human eye, microscopes, and telescopes.
Reflection Of Light By Spherical Mirrors
The behaviour of light when it bounces off a surface is governed by the laws of reflection. These laws state that:
1. The angle of incidence (the angle between the incoming ray and the perpendicular line to the surface, called the normal) is always equal to the angle of reflection (the angle between the outgoing ray and the normal).
2. The incident ray, the reflected ray, and the normal to the reflecting surface at the point where the light hits, all lie within the same plane.
These fundamental laws are applicable at every point on any reflecting surface, regardless of whether it is flat or curved. Our focus here will be on spherical mirrors, which are parts of a hollow sphere. For a spherical surface, the normal at any point is the line joining that point to the center of the sphere from which the mirror is cut. This line is equivalent to the radius at that point.
Key terms associated with spherical mirrors include:
- Pole (P): The geometrical center of the spherical mirror's reflecting surface.
- Centre of Curvature (C): The center of the imaginary sphere of which the mirror is a part.
- Principal Axis: The straight line passing through the pole and the center of curvature of the spherical mirror.
Sign Convention
To consistently apply mathematical formulas for mirrors and lenses, a standard system for measuring distances and heights is essential. We follow the Cartesian sign convention:
1. All distances are measured from the pole (P) of the mirror or the optical center of a lens.
2. Distances measured in the same direction as the incoming (incident) light are considered positive.
3. Distances measured in the direction opposite to the incoming (incident) light are considered negative.
4. Heights measured upwards from and perpendicular to the principal axis are considered positive.
5. Heights measured downwards from and perpendicular to the principal axis are considered negative.
Adopting this convention ensures that a single formula can describe image formation for various cases, including different types of mirrors/lenses and real/virtual images.
Focal Length Of Spherical Mirrors
When a beam of light rays traveling parallel to the principal axis strikes a spherical mirror, these rays either converge to a point or appear to diverge from a point after reflection. This point is called the principal focus (F).
- For a concave mirror, parallel paraxial rays converge to a real focus point on the principal axis in front of the mirror.
- For a convex mirror, parallel paraxial rays diverge such that they appear to originate from a virtual focus point located behind the mirror on the principal axis.
The focal plane is a plane passing through the principal focus and perpendicular to the principal axis. If parallel rays are incident at an angle to the principal axis, they converge to a point (or appear to diverge from a point) in the focal plane.
The distance between the pole (P) and the principal focus (F) is called the focal length (f) of the mirror. For spherical mirrors with a small aperture compared to their radius of curvature (paraxial approximation), the focal length is related to the radius of curvature ($R$) by the formula:
$$f = \frac{R}{2}$$
This relationship can be derived using geometric principles and the laws of reflection, assuming that the incident rays are paraxial (close to the principal axis and making small angles). In the paraxial approximation, trigonometric functions of small angles can be approximated by the angle itself ($\tan \theta \approx \theta$), and distances measured along the principal axis can be approximated by distances from the pole.
The Mirror Equation
An image is formed at a point where light rays emanating from an object point actually meet after reflection or refraction (real image) or appear to meet after being extended backwards (virtual image). Image formation establishes a correspondence between points on the object and points on the image.
To find the image of an object formed by a spherical mirror, we can trace the path of specific rays originating from the object. Convenient rays to trace include:
1. A ray parallel to the principal axis reflects through the principal focus (concave) or appears to come from the principal focus (convex).
2. A ray passing through the center of curvature (concave) or directed towards the center of curvature (convex) strikes the mirror normally and reflects back along its original path.
3. A ray passing through the principal focus (concave) or directed towards the principal focus (convex) reflects parallel to the principal axis.
4. A ray incident at the pole reflects according to the law of reflection, with the principal axis acting as the normal.
Tracing at least two of these rays from a point on the object helps locate the corresponding image point. The image of an extended object can be found by locating the images of several points on the object (e.g., the endpoints).
The mathematical relationship between the object distance ($u$), image distance ($v$), and focal length ($f$) for a spherical mirror is given by the mirror equation:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
Here, $u$ is the distance of the object from the pole, and $v$ is the distance of the image from the pole. Distances must be used with appropriate signs based on the Cartesian sign convention.
The linear magnification (m) quantifies how much the image is enlarged or reduced compared to the object. It is defined as the ratio of the height of the image ($h'$) to the height of the object ($h$):
$$m = \frac{h'}{h}$$
Using similar triangles formed by the object, image, and principal axis, the magnification can also be related to the image and object distances:
$$m = -\frac{v}{u}$$
The sign of $m$ indicates the image orientation: a positive $m$ signifies an erect image, while a negative $m$ indicates an inverted image. These equations ($1/v + 1/u = 1/f$ and $m = -v/u$) are valid for all types of spherical mirrors (concave and convex) and for both real and virtual images when the sign convention is applied correctly.
Example 9.1. Suppose that the lower half of the concave mirror’s reflecting surface in Fig. 9.6 is covered with an opaque (non-reflective) material. What effect will this have on the image of an object placed in front of the mirror?
Answer:
Even if a portion of the mirror is covered, the *entire* image of the object will still be formed. This is because every point on the object emits numerous rays of light, and many of these rays that strike the *uncovered* part of the mirror contribute to forming the image at the same location. However, since fewer rays are reflecting off the mirror surface and contributing to the image, the intensity or brightness of the image will be reduced. If half the mirror is covered, the image intensity will be approximately halved.
Example 9.2. A mobile phone lies along the principal axis of a concave mirror, as shown in Fig. 9.7. Show by suitable diagram, the formation of its image. Explain why the magnification is not uniform. Will the distortion of image depend on the location of the phone with respect to the mirror?
Answer:
For points on the mobile phone that are at different distances along the principal axis from the mirror, the image distance ($v$) calculated using the mirror formula will be different. The magnification ($m = -v/u$) also depends on both $u$ and $v$. Since $u$ varies for different parts of the phone lying along the axis, $v$ and consequently $m$ will also vary. This leads to non-uniform magnification for parts of the object located at different distances from the mirror along the principal axis. The end of the phone closer to the mirror will be magnified differently than the end farther away. This non-uniform magnification causes the image to be distorted. The degree of distortion depends on the range of object distances covered by the phone and the mirror's focal length, hence it depends on the phone's location relative to the mirror.
Example 9.3. An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.
Answer:
The radius of curvature $R = 15 \text{ cm}$. For a concave mirror, the focal length $f = R/2 = 15/2 = 7.5 \text{ cm}$. According to the sign convention, $R$ and $f$ for a concave mirror are negative, so $R = -15 \text{ cm}$ and $f = -7.5 \text{ cm}$.
The object is placed in front of the mirror, so the object distance $u$ is negative.
(i) Object distance $u = -10 \text{ cm}$. Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$$\frac{1}{v} + \frac{1}{-10} = \frac{1}{-7.5}$$
$$\frac{1}{v} = -\frac{1}{7.5} + \frac{1}{10} = \frac{-10 + 7.5}{7.5 \times 10} = \frac{-2.5}{75}$$
$$v = \frac{75}{-2.5} = -30 \text{ cm}$$
The image is formed at $30 \text{ cm}$ in front of the mirror (since $v$ is negative). As $v$ is negative and on the same side as the object, the image is real and inverted.
Magnification $m = -\frac{v}{u} = -\frac{-30}{-10} = -3$. The magnification is $|-3|=3$, meaning the image is 3 times larger than the object. The negative sign confirms it is inverted.
Nature: Real, Inverted, Magnified.
(ii) Object distance $u = -5 \text{ cm}$. Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$$\frac{1}{v} + \frac{1}{-5} = \frac{1}{-7.5}$$
$$\frac{1}{v} = -\frac{1}{7.5} + \frac{1}{5} = \frac{-5 + 7.5}{7.5 \times 5} = \frac{2.5}{37.5}$$
$$v = \frac{37.5}{2.5} = 15 \text{ cm}$$
The image is formed at $15 \text{ cm}$ behind the mirror (since $v$ is positive). As $v$ is positive and on the opposite side of the object, the image is virtual and erect.
Magnification $m = -\frac{v}{u} = -\frac{15}{-5} = 3$. The magnification is $|3|=3$, meaning the image is 3 times larger than the object. The positive sign confirms it is erect.
Nature: Virtual, Erect, Magnified.
Example 9.4. Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R = 2 m. If the jogger is running at a speed of 5 m s–1, how fast the image of the jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m, and (d) 9 m away.
Answer:
The side view mirror of a car is typically a convex mirror. The radius of curvature is given as $R = 2 \text{ m}$. For a convex mirror, $R$ is positive, so $R = +2 \text{ m}$. The focal length is $f = R/2 = +2/2 = +1 \text{ m}$.
The jogger is approaching the car, so the object distance $u$ is negative and decreasing. Let the jogger's position at time $t$ be $u(t)$ and the image position be $v(t)$. The speed of the jogger is $du/dt = -5 \text{ m/s}$ (negative because $u$ is decreasing). We want to find the speed of the image, $|dv/dt|$.
From the mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, we can express $v$ in terms of $u$ and $f$: $v = \frac{fu}{u-f}$.
To find the image speed, we differentiate $v$ with respect to time $t$ using the chain rule: $\frac{dv}{dt} = \frac{d}{dt}\left(\frac{fu}{u-f}\right)$.
Using the quotient rule $\frac{d}{dx}\left(\frac{g(x)}{h(x)}\right) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$:
Here, $g(u) = fu$ and $h(u) = u-f$. So $g'(u) = f$ and $h'(u) = 1$.
$$\frac{dv}{dt} = f \times \frac{(1)\times(u-f) - (u)\times(1)}{(u-f)^2} \times \frac{du}{dt}$$
$$\frac{dv}{dt} = f \times \frac{u-f - u}{(u-f)^2} \times \frac{du}{dt} = \frac{-f^2}{(u-f)^2} \frac{du}{dt}$$
Given $f = +1 \text{ m}$ and $du/dt = -5 \text{ m/s}$. So $\frac{dv}{dt} = \frac{-(+1)^2}{(u-(+1))^2} (-5) = \frac{-1}{(u-1)^2} (-5) = \frac{5}{(u-1)^2}$.
Since the jogger is approaching from in front of the mirror, $u$ is always negative. For example, $u = -39 \text{ m}$. So $u-1$ will be a larger negative number, and $(u-1)^2$ will be positive. The image speed is positive, meaning it moves away from the mirror as the jogger approaches, which is characteristic for a convex mirror.
Now we calculate the image speed $|dv/dt|$ for the given object distances:
(a) $u = -39 \text{ m}$: Speed $= \frac{5}{(-39 - 1)^2} = \frac{5}{(-40)^2} = \frac{5}{1600} = \frac{1}{320} \text{ m/s}$.
(b) $u = -29 \text{ m}$: Speed $= \frac{5}{(-29 - 1)^2} = \frac{5}{(-30)^2} = \frac{5}{900} = \frac{1}{180} \text{ m/s}$.
(c) $u = -19 \text{ m}$: Speed $= \frac{5}{(-19 - 1)^2} = \frac{5}{(-20)^2} = \frac{5}{400} = \frac{1}{80} \text{ m/s}$.
(d) $u = -9 \text{ m}$: Speed $= \frac{5}{(-9 - 1)^2} = \frac{5}{(-10)^2} = \frac{5}{100} = \frac{1}{20} \text{ m/s}$.
Comparing these speeds: $1/320 \approx 0.0031 \text{ m/s}$, $1/180 \approx 0.0056 \text{ m/s}$, $1/80 \approx 0.0125 \text{ m/s}$, $1/20 = 0.05 \text{ m/s}$.
The image speed increases significantly as the jogger gets closer to the mirror, even though the jogger's speed is constant. This demonstrates why objects in side-view mirrors appear to "speed up" as they get closer.
Refraction
When a beam of light transitions from one transparent medium into another, it undergoes a change in direction at the boundary separating the two media. This phenomenon is called refraction. A portion of the light is also reflected back into the original medium (this is called internal reflection when light goes from denser to rarer), but the bending of light as it enters the new medium is the defining characteristic of refraction.
For light incident obliquely (at an angle other than 0° or 90°), the direction of propagation changes. The fundamental rules governing refraction are known as Snell's laws of refraction, based on experimental observations:
1. The incident ray, the refracted ray, and the normal (the line perpendicular to the interface) at the point where the light hits, all lie in the same plane.
2. For a given pair of media and a specific colour (wavelength) of light, the ratio of the sine of the angle of incidence ($\sin i$) to the sine of the angle of refraction ($\sin r$) is a constant. This constant is called the refractive index of the second medium with respect to the first medium, denoted as $n_{21}$.
$$\frac{\sin i}{\sin r} = n_{21}$$
If $n_{21} > 1$, then $\sin i > \sin r$. Since the sine function increases with angle in the relevant range (0° to 90°), this implies $i > r$. In this case, the refracted ray bends towards the normal. Medium 2 is considered optically denser than medium 1.
If $n_{21} < 1$, then $\sin i < \sin r$, implying $i < r$. The refracted ray bends away from the normal. Medium 2 is considered optically rarer than medium 1.
It's important to distinguish between optical density and mass density. Optical density is related to how much a medium slows down light (inversely proportional to the speed of light in the medium), while mass density is mass per unit volume. A medium with higher mass density is not necessarily optically denser. For example, turpentine has lower mass density than water but is optically denser, causing light to bend more towards the normal when entering it from air compared to water.
The refractive index of medium 1 with respect to medium 2, $n_{12}$, is the reciprocal of the refractive index of medium 2 with respect to medium 1, $n_{21}$:
$$n_{12} = \frac{1}{n_{21}}$$
Refractive indices are relative, but often we refer to the absolute refractive index of a medium, which is its refractive index with respect to vacuum or air (which has a refractive index very close to 1). Let $n_1$ be the absolute refractive index of medium 1 and $n_2$ be the absolute refractive index of medium 2. Then $n_{21} = n_2/n_1$. Snell's law can be written as $n_1 \sin i = n_2 \sin r$. For a chain of media 1, 2, 3, the relationship between refractive indices is $n_{31} = n_{32} \times n_{21}$, where $n_{31} = n_3/n_1$, $n_{32} = n_3/n_2$, and $n_{21} = n_2/n_1$. This shows $(n_3/n_1) = (n_3/n_2) \times (n_2/n_1)$, which is consistent.
Simple consequences of refraction include the lateral shift of light passing through a parallel-sided glass slab (the emergent ray is parallel to the incident ray but displaced sideways) and the phenomenon of objects appearing shallower when viewed through a denser medium like water. For viewing nearly perpendicular to the surface, the apparent depth ($h_{apparent}$) is related to the real depth ($h_{real}$) and the refractive index ($n$) of the medium by:
$$h_{apparent} = \frac{h_{real}}{n}$$
Total Internal Reflection
When light travels from an optically denser medium to an optically rarer medium (e.g., from water to air), it undergoes refraction and bends away from the normal. Simultaneously, a portion of the light is reflected back into the denser medium; this is called internal reflection.
As the angle of incidence ($i$) in the denser medium increases, the angle of refraction ($r$) in the rarer medium also increases (since $r > i$). There comes a specific angle of incidence for which the refracted ray grazes the boundary between the two media, meaning the angle of refraction is 90°. This particular angle of incidence is known as the critical angle ($i_c$).
According to Snell's law, at the critical angle: $n_{denser} \sin i_c = n_{rarer} \sin 90^\circ$. Since $\sin 90^\circ = 1$, we get:
$$\sin i_c = \frac{n_{rarer}}{n_{denser}} = \frac{1}{n_{denser, rarer}}$$
where $n_{denser, rarer}$ is the refractive index of the denser medium with respect to the rarer medium ($n_{denser}/n_{rarer}$).
If the angle of incidence ($i$) in the denser medium becomes greater than the critical angle ($i > i_c$), refraction is no longer possible. All the light is reflected back into the denser medium. This phenomenon is called Total Internal Reflection (TIR). Unlike normal reflection where some light is always transmitted, in TIR, there is theoretically no loss of light due to transmission, resulting in a highly intense reflected ray.
Total Internal Reflection In Nature And Its Technelogical Applications
Total Internal Reflection is responsible for several fascinating natural phenomena and is extensively used in modern technology:
1. Mirage: An atmospheric optical phenomenon where light is refracted to produce a displaced image of distant objects or the sky. It occurs when light passes through layers of air with different temperatures and thus different refractive indices, leading to TIR, especially in hot conditions near the ground.
2. Brilliance of Diamonds: Diamonds sparkle brightly due to their very high refractive index (around 2.42) which results in a small critical angle (about 24.4°). When light enters a diamond, it undergoes multiple total internal reflections inside before emerging, giving it its characteristic brilliance.
3. Prisms: Right-angled prisms are designed with angles such that light entering one face undergoes TIR at another face, allowing them to deviate light by 90° or 180° without significant loss of intensity. They can also be used to invert images without changing their size. For this, the critical angle of the prism material must be less than 45°.
4. Optical Fibres: These are thin strands of high-quality glass or quartz that transmit light signals over long distances using TIR. An optical fibre consists of a core (higher refractive index) surrounded by cladding (lower refractive index). Light entering the core at a suitable angle strikes the core-cladding interface at an angle greater than the critical angle and undergoes repeated TIR along the length of the fibre, even when the fibre is bent. This mechanism ensures minimal signal loss. Optical fibres are crucial for high-speed internet, telecommunications, and are also used in medicine (endoscopes) and decorative lighting.
Here is a table showing critical angles for some substances with respect to air:
| Substance medium | Refractive index | Critical angle |
|---|---|---|
| Water | 1.33 | 48.75° |
| Crown glass | 1.52 | 41.14° |
| Dense flint glass | 1.62 | 37.31° |
| Diamond | 2.42 | 24.41° |
Refraction At Spherical Surfaces And By Lenses
Extending the principles of refraction from plane surfaces, we can analyze how light bends when it passes through a curved boundary between two transparent media. A small section of a spherical surface can be treated as planar, and Snell's laws apply at each point of incidence. For a spherical surface, the normal at any point is a line passing through the center of curvature of the sphere.
Thin lenses are optical components made of transparent material (like glass or plastic) bounded by two surfaces, at least one of which is spherical. Common types are biconvex, biconcave, plano-convex, plano-concave, etc. Image formation by a lens can be understood as the result of two successive refractions, one at each spherical surface.
Refraction At A Spherical Surface
Consider a spherical surface separating two media with refractive indices $n_1$ and $n_2$. Let the center of curvature be C and the radius of curvature be R. For paraxial rays originating from an object point O in medium 1 and forming an image point I in medium 2, the relationship between object distance ($u$, measured from the surface), image distance ($v$, measured from the surface), refractive indices ($n_1$, $n_2$), and radius of curvature ($R$) is given by the formula:
$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$
This formula holds for any spherical refracting surface (convex or concave) and applies with proper use of the Cartesian sign convention for distances $u$, $v$, and $R$. $u$ is typically negative for a real object in front of the surface (incident light direction is positive), $v$ is positive for a real image formed on the side of refracted light, and $R$ is positive for a convex surface (center of curvature on the side of refracted light) and negative for a concave surface (center of curvature on the side of incident light), following the standard convention relative to the pole.
Example 9.5. Light from a point source in air falls on a spherical glass surface (n = 1.5 and radius of curvature = 20 cm). The distance of the light source from the glass surface is 100 cm. At what position the image is formed?
Answer:
We use the spherical refracting surface formula: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.
Medium 1 is air, so $n_1 = 1$. Medium 2 is glass, so $n_2 = 1.5$.
The light source (object) is in air, $100 \text{ cm}$ from the surface. Since it's in front of the surface and light is incident from left to right (standard convention), the object distance $u = -100 \text{ cm}$.
The radius of curvature is $20 \text{ cm}$. The example doesn't specify if it's a convex or concave surface, but based on typical problem types and the positive result for $v$, it implies a convex surface, meaning the center of curvature is on the side where light refracts. So, $R = +20 \text{ cm}$.
We need to find the image distance $v$.
$$\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{+20}$$
$$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20} = \frac{5}{200} = \frac{1}{40}$$
$$\frac{1.5}{v} = \frac{1}{40} - \frac{1}{100} = \frac{5 - 2}{200} = \frac{3}{200}$$
$$1.5 v = \frac{200}{3}$$
$$v = \frac{200}{3 \times 1.5} = \frac{200}{4.5} = \frac{2000}{45} = \frac{400}{9} \approx 44.44 \text{ cm}$$
Looking back at the provided solution in the text, it seems there was a numerical error or a different interpretation in the original source. The given solution of +100 cm is incorrect based on the formula and inputs provided ($u = -100, R = +20, n_1=1, n_2=1.5$). Let's double-check the provided example answer's arithmetic:
Original text's calculation steps:
$$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}$$
$$\frac{1.5}{v} = \frac{0.5}{20} - \frac{1}{100} = \frac{5}{200} - \frac{2}{200} = \frac{3}{200}$$
$$v = \frac{1.5 \times 200}{3} = \frac{300}{3} = 100 \text{ cm}$$
Ah, my calculation error was in the final step. The textbook's calculation is correct assuming $u=-100$, $R=+20$, $n_1=1$, $n_2=1.5$.
So, the correct answer based on the provided text and calculation:
$$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}$$
$$\frac{1.5}{v} = \frac{0.5}{20} - \frac{1}{100} = \frac{1}{40} - \frac{1}{100} = \frac{5 - 2}{200} = \frac{3}{200}$$
$$v = \frac{1.5 \times 200}{3} = \frac{300}{3} = 100 \text{ cm}$$
The image distance is $v = +100 \text{ cm}$. Since $v$ is positive, the image is formed on the side where the refracted light travels, which is behind the glass surface at a distance of $100 \text{ cm}$. This is a real image.
Refraction By A Lens
Image formation by a thin lens (a lens where the thickness is negligible compared to object/image distances and radii of curvature) can be analyzed by considering refraction at each of its two surfaces sequentially. Light from an object first refracts at the front surface, forming an intermediate image. This intermediate image then acts as the object (either real or virtual) for the second surface, which performs another refraction to form the final image.
Applying the spherical surface refraction formula $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$ to each surface of a thin lens and combining the results leads to the lens maker's formula:
$$\frac{1}{f} = \left(\frac{n_{lens}}{n_{medium}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
Here, $f$ is the focal length of the lens, $n_{lens}$ is the refractive index of the lens material, $n_{medium}$ is the refractive index of the surrounding medium, $R_1$ is the radius of curvature of the first surface the light encounters, and $R_2$ is the radius of curvature of the second surface. $R_1$ and $R_2$ are used with their proper signs according to the convention ($R$ is positive if the center of curvature is on the side of light travel after refraction from that surface, and negative otherwise).
The focal length $f$ is a property of the lens and the medium it's in. It's positive for a converging lens (like a convex lens in air) and negative for a diverging lens (like a concave lens in air). The lens maker's formula is useful for designing lenses with desired focal lengths.
For a thin lens, the relationship between the object distance ($u$), image distance ($v$), and focal length ($f$) is given by the thin lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
This formula is universally applicable to all thin lenses (convex and concave) and all types of images (real and virtual) when distances are substituted with appropriate signs. The object distance $u$ is measured from the optical center (the central point of the thin lens), and the image distance $v$ is also measured from the optical center.
Similar to mirrors, specific rays can be traced to determine the image location and characteristics for lenses:
1. A ray parallel to the principal axis passes through the second principal focus F' after refraction (convex lens) or appears to diverge from the first principal focus F (concave lens).
2. A ray passing through the optical center (the geometric center of the thin lens) travels undeviated.
3. A ray passing through the first principal focus F (convex lens) or directed towards the second principal focus F' (concave lens) emerges parallel to the principal axis after refraction.
The magnification (m) produced by a lens is defined as the ratio of image height ($h'$) to object height ($h$). For a thin lens, this is related to image and object distances by:
$$m = \frac{h'}{h} = \frac{v}{u}$$
The sign of $m$ indicates the image orientation: positive for erect, negative for inverted.
Power Of A Lens
The ability of a lens to converge or diverge light rays incident upon it is quantified by its power (P). A lens with a shorter focal length bends light more strongly and thus has greater power.
Power is defined as the reciprocal of the focal length, provided the focal length is measured in meters:
$$P = \frac{1}{f}$$
The standard unit for lens power is the dioptre (D). One dioptre is the power of a lens with a focal length of 1 meter ($1 D = 1 \text{ m}^{-1}$).
Converging lenses (like convex lenses in air) have a positive focal length and therefore positive power. Diverging lenses (like concave lenses in air) have a negative focal length and thus negative power.
For instance, a prescription of +2.5 D indicates a convex lens with a focal length of $1/2.5 \text{ m} = 0.4 \text{ m} = 40 \text{ cm}$. A prescription of -4.0 D indicates a concave lens with a focal length of $1/(-4.0) \text{ m} = -0.25 \text{ m} = -25 \text{ cm}$.
Example 9.7. (i) If f = 0.5 m for a glass lens, what is the power of the lens? (ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass? (iii) A convex lens has 20 cm focal length in air. What is focal length in water? (Refractive index of air-water = 1.33, refractive index for air-glass = 1.5.)
Answer:
(i) Focal length $f = 0.5 \text{ m}$. The power of the lens is $P = 1/f = 1/0.5 = 2 \text{ D}$.
(ii) This is a double convex lens. Let's assume light is incident from the left. The first surface is convex towards the incident light, so its radius of curvature $R_1 = +10 \text{ cm}$. The second surface is convex towards the right, so its radius of curvature $R_2 = -15 \text{ cm}$ (the center of curvature is on the left, opposite to the refracted light direction from the first surface). The focal length in air is $f = 12 \text{ cm}$. We use the lens maker's formula: $\frac{1}{f} = \left(\frac{n_{lens}}{n_{medium}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
Here, $n_{medium}$ is the refractive index of air, which is approximately 1. Let the refractive index of the glass be $n$.
$$\frac{1}{12 \text{ cm}} = (n - 1) \left(\frac{1}{+10 \text{ cm}} - \frac{1}{-15 \text{ cm}}\right)$$
$$\frac{1}{12} = (n - 1) \left(\frac{1}{10} + \frac{1}{15}\right) = (n - 1) \left(\frac{3 + 2}{30}\right) = (n - 1) \left(\frac{5}{30}\right) = (n - 1) \left(\frac{1}{6}\right)$$
$$n - 1 = \frac{6}{12} = \frac{1}{2} = 0.5$$
$$n = 1 + 0.5 = 1.5$$
The refractive index of the glass is 1.5.
(iii) The convex lens has focal length $f_{air} = 20 \text{ cm}$ in air. Refractive index of air ($n_{air}$) $\approx 1$. Refractive index of glass ($n_{glass}$) is given as 1.5. Refractive index of water ($n_{water}$) is 1.33.
Using the lens maker's formula for the lens in air:
$$\frac{1}{f_{air}} = \left(\frac{n_{glass}}{n_{air}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
$$\frac{1}{20} = \left(\frac{1.5}{1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (0.5) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad (*)$$
Now, the same lens is placed in water. The radii of curvature $R_1$ and $R_2$ remain the same. Let the focal length in water be $f_{water}$.
$$\frac{1}{f_{water}} = \left(\frac{n_{glass}}{n_{water}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
$$\frac{1}{f_{water}} = \left(\frac{1.5}{1.33} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad (**)$$
From equation (*), we have $\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{20 \times 0.5} = \frac{1}{10}$. Substitute this into equation (**):
$$\frac{1}{f_{water}} = \left(\frac{1.5 - 1.33}{1.33}\right) \left(\frac{1}{10}\right) = \left(\frac{0.17}{1.33}\right) \left(\frac{1}{10}\right)$$
$$f_{water} = \frac{1.33 \times 10}{0.17} = \frac{13.3}{0.17} \approx 78.24 \text{ cm}$$
The focal length of the convex lens in water is approximately +78.2 cm. Notice that the focal length increases when the lens is placed in a denser medium, and the power ($P = 1/f_{water}$) decreases.
Combination Of Thin Lenses In Contact
When two or more thin lenses are placed in contact with each other such that their principal axes coincide, they form a lens system. The effect of this combination can often be represented by a single equivalent lens with a specific focal length and power.
Consider two thin lenses A and B with focal lengths $f_1$ and $f_2$ in contact. An object O is placed in front of the first lens A. Lens A forms an image I$_1$. If lens B were not present, this image would be formed at a distance $v_1$ from the lenses. Using the thin lens formula for lens A: $\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1}$.
This intermediate image I$_1$ now serves as the object for the second lens B. The object distance for lens B is $u_2 = v_1$. If the intermediate image is formed beyond lens B, it is a virtual object for lens B (positive object distance in this convention). If it's formed before or at lens B, it's a real object (negative object distance). The second lens B forms the final image I at a distance $v$ from the lenses. Using the thin lens formula for lens B: $\frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2}$.
Adding the equations for lens A and lens B:
$$\left(\frac{1}{v_1} - \frac{1}{u}\right) + \left(\frac{1}{v} - \frac{1}{v_1}\right) = \frac{1}{f_1} + \frac{1}{f_2}$$
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2}$$
If we consider the combination as a single equivalent thin lens with focal length $f_{eq}$, its lens formula is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{eq}}$. Comparing the equations, we get the formula for the effective focal length of two thin lenses in contact:
$$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$$
For a combination of several thin lenses in contact with focal lengths $f_1, f_2, f_3, \dots$, the effective focal length is given by:
$$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} + \dots$$
In terms of power, since $P = 1/f$, the total power of the combination is the algebraic sum of the individual lens powers:
$$P_{total} = P_1 + P_2 + P_3 + \dots$$
This addition is algebraic, meaning powers of convex lenses are positive and powers of concave lenses are negative. Combining lenses allows for achieving desired magnification, image orientation, and correcting optical defects (aberrations).
The total magnification produced by a combination of lenses is the product of the magnifications produced by each lens individually:
$$m_{total} = m_1 \times m_2 \times m_3 \times \dots$$
Example 9.8. Find the position of the image formed by the lens combination given in the Fig. 9.20.
*(Note: Fig 9.20 shows three lenses along an axis. Lens 1 (Convex, f=10cm) at origin, Object 30cm to its left. Lens 2 (Concave, f=-10cm) 5cm to the right of Lens 1. Lens 3 (Convex, f=30cm) 10cm to the right of Lens 2.)*
Answer:
We will find the image formed by each lens sequentially, using the image of one lens as the object for the next.
For the first lens (Convex, $f_1 = +10 \text{ cm}$):
Object distance $u_1 = -30 \text{ cm}$ (object is 30cm to the left, assuming light from left to right is positive).
Using the thin lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:
$$\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{10}$$
$$\frac{1}{v_1} + \frac{1}{30} = \frac{1}{10}$$
$$\frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{3 - 1}{30} = \frac{2}{30} = \frac{1}{15}$$
$$v_1 = +15 \text{ cm}$$
The image formed by the first lens is $15 \text{ cm}$ to the right of the first lens. This is a real image.
For the second lens (Concave, $f_2 = -10 \text{ cm}$):
The first image is at $+15 \text{ cm}$ from the first lens. The second lens is $5 \text{ cm}$ to the right of the first lens. So, the distance of the first image from the second lens is $u_2 = v_1 - (\text{distance between lenses}) = 15 \text{ cm} - 5 \text{ cm} = 10 \text{ cm}$.
Since this intermediate image is to the right of the second lens (in the direction of light), it acts as a virtual object for the second lens. Thus, $u_2 = +10 \text{ cm}$.
Using the thin lens formula for the second lens $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:
$$\frac{1}{v_2} - \frac{1}{+10} = \frac{1}{-10}$$
$$\frac{1}{v_2} = -\frac{1}{10} + \frac{1}{10} = 0$$
$$v_2 = \infty$$
The image formed by the second lens is at infinity.
For the third lens (Convex, $f_3 = +30 \text{ cm}$):
The image formed by the second lens is at infinity. An object at infinity for a lens forms an image at the focal point. The third lens is $10 \text{ cm}$ to the right of the second lens.
The image from the second lens (at infinity) acts as the object for the third lens. So, the object distance $u_3 = \infty$.
Using the thin lens formula for the third lens $\frac{1}{v_3} - \frac{1}{u_3} = \frac{1}{f_3}$:
$$\frac{1}{v_3} - \frac{1}{\infty} = \frac{1}{30}$$
$$\frac{1}{v_3} - 0 = \frac{1}{30}$$
$$v_3 = +30 \text{ cm}$$
The final image is formed at $+30 \text{ cm}$ from the third lens. Since $v_3$ is positive, it is formed $30 \text{ cm}$ to the right of the third lens.
Refraction Through A Prism
A prism is a transparent optical element with flat, polished surfaces that refract light. Typically, it has a triangular cross-section with two refracting surfaces inclined at an angle to each other, called the angle of the prism (A). When a ray of light passes through a prism, it undergoes two refractions, one at each face.
Consider a ray incident on one face at an angle $i$. It refracts into the prism at an angle $r_1$. Inside the prism, it travels to the second face and is incident upon it at an angle $r_2$. It then refracts out of the prism at an angle $e$ (angle of emergence).
The total change in the direction of the ray as it passes through the prism is called the angle of deviation ($\delta$). It is the angle between the emergent ray and the direction of the incident ray. The angles involved are related by the following geometric principles:
1. The angle of the prism ($A$) is related to the angles of refraction inside the prism: $A = r_1 + r_2$.
2. The total deviation ($\delta$) is the sum of the deviations at each surface: $\delta = (i - r_1) + (e - r_2) = i + e - (r_1 + r_2)$. Substituting the first relation, we get $\delta = i + e - A$.
The angle of deviation depends on the angle of incidence, the angle of the prism, and the refractive index of the prism material. As the angle of incidence varies, the deviation angle changes, reaching a minimum value called the angle of minimum deviation ($D_m$).
At minimum deviation, the following conditions hold:
1. The angle of incidence equals the angle of emergence ($i = e$).
2. The angles of refraction inside the prism are equal ($r_1 = r_2$).
From $A = r_1 + r_2$ and $r_1 = r_2$, we get $A = 2r$, so $r = A/2$.
From $\delta = i + e - A$ and $\delta = D_m$, $i = e$, we get $D_m = 2i - A$, so $i = (A + D_m)/2$.
Applying Snell's law ($n_1 \sin i = n_2 \sin r$) at the first face (assuming medium 1 is air with $n_1 \approx 1$ and medium 2 is prism material with refractive index $n$):
$$1 \times \sin i = n \sin r_1$$
Substituting the expressions for $i$ and $r$ at minimum deviation:
$$n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$
This formula is used to determine the refractive index of the prism material by measuring the angle of the prism ($A$) and the angle of minimum deviation ($D_m$).
For a thin prism (a prism with a very small angle $A$), the angle of minimum deviation $D_m$ is also small. Using the small angle approximation for sine ($\sin \theta \approx \theta$), the formula simplifies to:
$$n \approx \frac{(A + D_m)/2}{A/2} = \frac{A + D_m}{A} = 1 + \frac{D_m}{A}$$
Rearranging, the deviation for a thin prism is approximately:
$$\delta \approx (n - 1)A$$
Thin prisms produce small deviations and do not cause significant dispersion (splitting of white light into colours) compared to thicker prisms.
Optical Instruments
Various optical devices and instruments have been developed to extend or improve the capabilities of human vision, utilizing the principles of reflection and refraction. These include simple tools like periscopes and kaleidoscopes, as well as complex instruments like binoculars, telescopes, and microscopes. While the human eye itself is a sophisticated natural optical instrument, we will focus on the working principles of the microscope and the telescope here.
The Microscope
Microscopes are used to obtain magnified images of small objects. A basic type is the simple microscope, which is essentially a single converging lens (a convex lens) with a short focal length. To use it as a magnifier, the object is placed within or very close to the focal point of the lens. The lens forms a virtual, erect, and magnified image.
The image can be formed either at the near point (D), which is the least distance of distinct vision (approximately 25 cm for a normal eye), or at infinity (for relaxed viewing).
The magnification (m) for a simple microscope is typically referred to as angular magnification or magnifying power. It's the ratio of the angle subtended by the image at the eye to the angle subtended by the object if it were placed at the near point D and viewed directly.
When the final image is formed at the near point (D), the magnifying power is:
$$m = 1 + \frac{D}{f}$$
When the final image is formed at infinity (relaxed eye viewing), the magnifying power is:
$$m = \frac{D}{f}$$
Here, $f$ is the focal length of the lens and $D \approx 25 \text{ cm}$. A simple microscope provides a limited magnification, typically up to about 9x.
For achieving much higher magnifications, a compound microscope is used. It consists of two converging lenses:
1. The Objective Lens: Placed close to the object, it has a very short focal length ($f_o$). It forms a real, inverted, and magnified intermediate image.
2. The Eyepiece (or Ocular): Acts like a simple microscope, magnifying the intermediate image formed by the objective. It also has a short focal length ($f_e$). It forms the final, enlarged, virtual image. This final image is inverted with respect to the original object.
The intermediate image is usually positioned near or within the focal plane of the eyepiece. The distance between the second focal point of the objective and the first focal point of the eyepiece is called the tube length (L).
The total magnifying power ($m$) of a compound microscope is the product of the linear magnification produced by the objective ($m_o$) and the angular magnification produced by the eyepiece ($m_e$):
$$m = m_o \times m_e$$
The linear magnification of the objective is approximately $m_o \approx \frac{\text{distance of intermediate image from objective}}{\text{object distance from objective}}$. For the intermediate image formed near the eyepiece's focus, this is approximately $m_o \approx L/f_o$.
The angular magnification of the eyepiece is like that of a simple microscope:
- If the final image is at the near point: $m_e = 1 + D/f_e$
- If the final image is at infinity: $m_e = D/f_e$
Thus, for the final image at infinity, the total magnifying power is approximately:
$$m \approx \left(\frac{L}{f_o}\right) \left(\frac{D}{f_e}\right)$$
To achieve high magnification, both the objective ($f_o$) and the eyepiece ($f_e$) should have small focal lengths. Modern microscopes use multiple lens elements in both the objective and eyepiece to minimize optical aberrations and improve image quality.
Telescope
Telescopes are used to view distant objects, providing angular magnification so that these objects appear closer and larger. Similar to a compound microscope, a telescope also has two primary lenses:
1. The Objective Lens: This lens is placed towards the distant object and has a large focal length ($f_o$) and a large aperture (diameter). It collects light from the distant object and forms a real, inverted, and diminished intermediate image at its focal point.
2. The Eyepiece: This lens is placed near the observer's eye and has a short focal length ($f_e$). It acts as a magnifier, magnifying the intermediate image formed by the objective. The final image formed by the eyepiece is typically virtual and inverted with respect to the original object.
For viewing distant objects, the intermediate image is formed at the focal point of the objective. In normal adjustment (when the final image is formed at infinity for relaxed viewing), this intermediate image is also located at the focal point of the eyepiece. The distance between the objective and the eyepiece in this case is $L = f_o + f_e$.
The magnifying power (m) of a telescope is defined as the ratio of the angle ($\beta$) subtended by the final image at the eye to the angle ($\alpha$) subtended by the distant object at the objective (which is approximately the same as the angle subtended at the eye):
$$m = \frac{\beta}{\alpha} = \frac{f_o}{f_e}$$
A telescope with a larger focal length objective and a smaller focal length eyepiece will provide higher magnifying power.
Key considerations for astronomical telescopes are:
1. Light Gathering Power: This determines how faint an object can be observed and depends on the area of the objective lens (proportional to the square of its diameter). Larger objectives collect more light.
2. Resolving Power: This is the ability to distinguish between two closely spaced distant objects. It depends on the diameter of the objective and the wavelength of light.
Therefore, astronomical telescopes ideally have objectives with large diameters.
While refracting telescopes use lenses, large lenses are difficult and expensive to manufacture, support (they can only be supported at the edges), and they suffer from chromatic aberration (different colours focus at slightly different points). For these reasons, modern large telescopes are often reflecting telescopes, which use concave mirrors as objectives.
Advantages of reflecting telescopes:
- No chromatic aberration (mirrors reflect all wavelengths equally).
- Easier mechanical support (mirrors can be supported from the back).
- Can be made with very large apertures.
One challenge is that the image is formed inside the telescope tube, potentially requiring the observer or a secondary mirror to be placed in the light path. The Cassegrain telescope design uses a secondary convex mirror to reflect the light through a hole in the primary concave mirror, allowing the eyepiece to be placed behind the main mirror.
Exercises
Question 9.1. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Answer:
Question 9.2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Answer:
Question 9.3. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Answer:
Question 9.4. Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)].
Answer:
Question 9.5. A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Answer:
Question 9.6. A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Answer:
Question 9.7. Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?
Answer:
Question 9.8. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is
(a) a convex lens of focal length 20cm, and
(b) a concave lens of focal length 16cm?
Answer:
Question 9.9. An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Answer:
Question 9.10. What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length 20cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Answer:
Question 9.11. A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at
(a) the least distance of distinct vision (25cm), and
(b) at infinity? What is the magnifying power of the microscope in each case?
Answer:
Question 9.12. A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,
Answer:
Question 9.13. A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Answer:
Question 9.14. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^6\text{m}$, and the radius of lunar orbit is $3.8 \times 10^8\text{m}$.
Answer:
Question 9.15. Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
Answer:
Question 9.16. A small pin fixed on a table top is viewed from above from a distance of 50cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
Answer:
Question 9.17. (a) Figure 9.28 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure. (b) What is the answer if there is no outer covering of the pipe?
Answer:
Question 9.18. The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
Answer:
Question 9.19. A screen is placed 90cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20cm. Determine the focal length of the lens.
Answer:
Question 9.20. (a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all? (b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40cm. Determine the magnification produced by the two-lens system, and the size of the image.
Answer:
Question 9.21. At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
Answer:
Question 9.22. A card sheet divided into squares each of size $1 \text{ mm}^2$ is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. (a) What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b) What is the angular magnification (magnifying power) of the lens? (c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Answer:
Question 9.23. (a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power? (b) What is the magnification in this case? (c) Is the magnification equal to the magnifying power in this case? Explain.
Answer:
Question 9.24. What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of $6.25 \text{ mm}^2$. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
[Note: Exercises 9.22 to 9.24 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
Answer:
Question 9.25. Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Answer:
Question 9.26. An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound microscope?
Answer:
Question 9.27. A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25cm)?
Answer:
Question 9.28. (a) For the telescope described in Exercise 9.27 (a), what is the separation between the objective lens and the eyepiece? (b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens? (c) What is the height of the final image of the tower if it is formed at 25cm?
Answer:
Question 9.29. A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?
Answer:
Question 9.30. Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.29. A current in the coil produces a deflection of $3.5^\circ$ of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?
Answer:
Question 9.31. Figure 9.30 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid?
Answer: